Prove for any ideal $J$, $\operatorname{rad} J$ is an ideal of $A$, if $J$ is ideal of $A$.

Question

Let $A$ be a ring, and let $J$, $K$ be ideals of $A$. The radical of $J$ is the set $$\operatorname{rad}J=\{a \in A:a^n \in J \text{ for some } n \in \mathbb{Z}\}.$$ Prove for any ideal $J$, $\operatorname{rad}J$ is an ideal of $A$.

I know that I will start the proof supposing that $a^n \in J$ and $b^m \in J$, and will work to show $(a+b)^{m+n} \in J.$ How would I go about this? and how do I use this to prove the ideal?

Is $A$ a commutative ring? I don't believe this is true for general rings. — Aweygan, Oct 25 '16 at 02:48
Yes A is commutative, your right the property doesn't hold for all rings. — Ddbb1994, Oct 25 '16 at 02:50

Eugene Zhang · Answer 1 · 2016-10-25T18:58:53.723

0

Hint: Use binomial theorem to expand $(a+b)^{m+n}$ and prove that each $a^{m+n-k}b^k\in J$. Since $J$ is an ideal, the sum is in $J$.

edited Oct 25 '16 at 18:58

answered Oct 25 '16 at 02:40

Eugene Zhang

16,805

Prove for any ideal $J$, $\operatorname{rad} J$ is an ideal of $A$, if $J$ is ideal of $A$.

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